Question

Integrate (((2y+1)/4) + y )^8 ?
see comments for this written better.

Answer 1

\[\left( \frac{ (2y+1) }{ 4 }+y \right)^{9}\]

Answer 2

with respect to y

Answer 3

change it to:
\[\int\limits_{}^{}\left( \frac{ 6y + 1 }{ 4 } \right)^n dy\]

Answer 4

another way is to substitute the quantity inside the bracket as 'u'
then find du=...

Answer 5

ya, do that. i think i made a mistake ^_^

Answer 6

let ((6y + 1)/4) = u

Answer 7

I hate to sound dumb but how did you get from ((2y+1)/4) +y to (6y+1)/4?

Answer 8

add the fractions. they had to have same denominator so:

\[\frac{ 2y + 1 }{ 4 } + y = \frac{ 2y+1 }{ 4 } + y*\frac{ 4 }{4 } = \frac{ 2y + 1 }{ 4 } + \frac{ 4y }{ 4 } = \frac{ 6y+1 }{ 4 }\]

Answer 9

why do i feel that is not your initial question ?
if so whats you main question ?

Answer 10

of course (slapping forehead)

my initial question i had a typo.

Answer 11

hehe :P